top view binary tree added

Author: neha030

Data Structures/Trees/top_view_binary_tree.cpp

@@ -0,0 +1,132 @@
+/*
+Top view of binary tree
+Given a binary tree, print the nodes in left to right manner as visible from above the tree.
+Input is given in order of root left child then right child.
+For e.g. 1 2 3 4 5 6 -1 -1 -1 -1 -1 -1 -1
+Tree looks like 
+              1
+          /      \
+      2           3
+    /     \       /
+  4       5     6
+   
+nodes '5' and '6' will be overlapped by '1' as seen from above, 
+so when viewed from the top , we would see the nodes 4, 2, 1 and 3.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// class for defining the the node, left and right child pointer along 
+// with a constructor for assigning the "key" to the node.
+class node
+{
+public:
+    int val;
+    node *left;
+    node *right;
+
+    node(int d)
+    {
+        val = d;
+        left = NULL, right = NULL;
+    }
+};
+
+// function for taking the input from user and assigning the node values accordingly.
+// Queue has been used to assign the "keys" to left and right child alternatively
+node *maketree()
+{
+    int d;
+    queue<node *> q;
+    cout << "Enter the root value: ";
+    cin >> d;
+    node *root = new node(d);
+    q.push(root);
+    while (!q.empty())
+    {
+        node *temp = q.front();
+        q.pop();
+        int l, r;
+        cout << "\nEnter the left and right childs : ";
+        cin >> l >> r;
+        if (l != -1)
+        {
+            // assigning value to the left child pointer and inserting  is as the parent node to repeat the process
+            temp->left = new node(l);
+            q.push(temp->left);
+        }
+        if (r != -1)
+        {
+            // assigning value to the right child pointer and inserting it as the parent node to repeat the process
+            temp->right = new node(r);
+            q.push(temp->right);
+        }
+    }
+    cout << "\n";
+    return root;
+}
+
+// The Concept is combining the left side view and right side view of that tree, will be 
+// the in total, the top view as all overlapping will be overcomed.
+
+/* Now for printing the tree, firstly the left child are printed recursively.
+-->                     1
+                    /      \
+-->              2           3
+                /     \      /
+-->            4       5    6
+so the nodes 4, 2 and 1 are printed respectively
+*/
+void printleft(node *root)
+{
+    if (root == NULL)
+        return;
+    printleft(root->left);
+    cout << root->val << " ";
+    return;
+}
+
+/* Now for printing the tree, secondly the right child are printed recursively.
+                       1            
+                    /      \
+                 2           3      <--
+                /     \      /
+               4       5    6   
+so the node 3 is printed respectively. 
+*/
+void printright(node *root)
+{
+    if (root == NULL)
+        return;
+    cout << root->val << " ";
+    printright(root->right);
+    return;
+}
+
+int main()
+{
+    // calling the function for making the binary tree
+    node *root = maketree();
+    // calling the printleft and printright function for printing nodes and overall, the top view
+    cout << "Top view: ";
+    printleft(root);
+    printright(root->right);
+    return 0;
+}
+
+/*
+Sample Input:
+Enter the root value: 1
+Enter the left and right childs : 2 3
+Enter the left and right childs : 4 5
+Enter the left and right childs : 6 -1
+Enter the left and right childs : -1 -1
+Enter the left and right childs : -1 -1
+Enter the left and right childs : -1 -1
+Sample Output:
+Top view: 4 2 1 3 
+
+Time Complexity:  O(n), for traversing the nodes
+Space Complexity: O(n), since queue has been used for n nodes
+*/